If X = {4n - 3n - 1, n N} and Y = {9(n - 1) : n N}, where N is the set of natural numbers, then X ? Y is equal to

Question

If $X = {4^{n} - 3 n - 1 : n \in N}$ and $Y = {9 (n - 1) : n \in N}$ , where $N$ is the set of natural numbers, then $X \cup Y$ is equal to

Smart Achievers · Accepted Answer

(3)

$X = {4^{n} - 3 n - 1},$ $Y = {9 (n - 1) : n \in N}$

$4^{n} - 3 n - 1$ $= {(1 + 3)}^{n} - 3 n - 1$

$= (1 + {}^{n}{C_{1}} 3 + {}^{n}{C_{2}} 3^{2} + \dots + 3^{n}) - 3 n - 1$

$= (1 + 3 n) + 9 ({}^{n}C_{2} + \dots + 3^{n - 2}) - 3 n - 1$

$= 9 ({}^{n}C_{2} + \dots + 3^{n - 2}),$ $n \geq 2 = 9 k$

Also for $n = 1$ , $4^{n} - 3 n - 1 = 0,$ a multiple of 9

Every element in $X$ is a multiple of 9. But $Y$ contains all multiples of 9. Hence $X \cup Y = Y .$

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